VCE Methods Question Thread!

[keltingmeith]

With the probability notation for normal distribution, is it  X ~ N(u, o^2) or X ~ N(u,o) ?? With u = mean, o = s.d and o^2 = variance

The former - we always reference things in terms of their variance, not standard deviation. Variance is good, standard deviation is bad.

[TheCommando]

How do i do 26b and c
https://ibb.co/nKmmZa

[TheCommando]

http://imgur.com/a/pZMj7
For the answer of question 5
Why do they do +4y? And how do they know to do it

[peanut]

http://imgur.com/a/pZMj7
For the answer of question 5
Why do they do +4y? And how do they know to do it

y is a random letter they gave for the height. From the diagram, it's clear that the block has four edges which are heights. The equation 4x + 4(3x) + 4y = 20 is just an equation for the statement "The sum of the lengths of all its edges is 20cm."

[TheCommando]

y is a random letter they gave for the height. From the diagram, it's clear that the block has four edges which are heights. The equation 4x + 4(3x) + 4y = 20 is just an equation for the statement "The sum of the lengths of all its edges is 20cm."

So the 4(3x)
Is the bottom base and the 4y is that height of thoose 4 lengths of the big rectangle side

[Gogo14]

So the 4(3x)
Is the bottom base and the 4y is that height of thoose 4 lengths of the big rectangle side

Yep, y is the height and 4y is the sum of all the heights.

How do i do 26b and c
https://ibb.co/nKmmZa

Diff the equation, and use the trig identity sin^2+cos^2=1 to make every term as sine. Simplify and solve. If you cannot solve it directly, try subbing a=sin^2 and  use the quadratic formula.

[ringring]

Would appreciate some help with these two questions (attached below)

[keltingmeith]

Would appreciate some help with these two questions (attached below)

Have you read this guide on finding general solutions? Anything from it you were unsure about?

[TheCommando]

Diff the equation, and use the trig identity sin^2+cos^2=1 to make every term as sine. Simplify and solve. If you cannot solve it directly, try subbing a=sin^2 and  use the quadratic formula.

I dont get it sorry

[Wota]

Hi a few CASrelated questions here:

1) how to find both local and global minima and maxima using the CAS
2) how to do a linear approximation using the CAS
3) how to find inverse function using the CAS
4) how to type a hybrid function using the CAS

cheers!

[Sirius]

I dont get it sorry

Expanding on what Gogo14 already said....
You want to differentiate the equation first.
Then use the identity Cos²(x) + Sin²(x)=1 and rearrange it to get Cos²(x) = 1 - sin²(x).
Sub this into the equation which should give you a hidden quadratic. If you are still stuck check the attached working.
Note: I haven't completed the entire question. Hopefully you can take it from the quadratic. 
Let me know if it works 

[MisterNeo]

I dont get it sorry

How do i do 26b and c
https://ibb.co/nKmmZa

For Question b:
Differentiate the function, after you change 2sin(x)cos(x) into sin(2x) to make it easier. (Double angle formula. Might be too high level for this maths)



Stationary points at f'(x)=0...


When does sinx-cos2x=0? THERE ARE 3 STATIONARY POINTS
1

(Refer to the special triangles for this one.)
We just established that at π/6, sinx-cos2x=0.
2





(To find this one, you need the unity circle properties.)
3




But the question says x-values from 0 to 2x...
This is where the odd even properties of sin/cos come into play (look above).
The graph repeats itself, and there is a stationary point right before 2x, so you add a full cycle (2π) onto -π/2.

So...the stationary points at x=(π/6, 5π/6, and 3π/2) from 0<x<2x.

It's really confusing, but graphing it on Desmos does help.
AND you should memorise the trig identities if they are not on a formula sheet in the exam.

Hope this makes sense..

[Alein]

Hey, I was wondering where can I get the slides for the free lecture of maths methods that took place on Saturday at RMIT?

[Ahmad_A_1999]

Hey, could someone please help me with this 

http://imgur.com/a/Bkrr4

I'm stuck on part 'e'

Any help would be great appreciated!

[Peanut Butter]

Hey, could someone please help me with this 

http://imgur.com/a/Bkrr4

I'm stuck on part 'e'

Any help would be great appreciated!

Let x1 be the x-coordinate of the left side of the rectangle, and let x2 be the x-coordinate of the right side of the rectangle.

We want to find the length of the rectangle = x2 - x1

We know that the base of the triangle is 6 units (8-2 from the diagram). We also assume that the rectangle is in the centre of the triangle, meaning that there is x1 units between the corner of the triangle and the corner of the rectangle. Since it is in the middle, there is x1 units on both sides = 2x1 units altogether.

Therefore, our length = 6 - 2x1 (does this part make sense? This is the trickiest part of the question )

Since our length is in terms of x1, we sub x1 into our linear equation (from part a) to find the corresponding y-value for x1.

Y = 3x1 - 4

We then need to further subtract 2 units, as the rectangle is 2 units above the x-axis, to find the width.

Width = (3x1 - 4) - 2 = 3x1 - 6

Area = length x width
         = (6 - 2x1)(3x1 - 6)

Find the derivate of the area (in terms of x1). Make derivative = 0 to find maximum x1 values. Sub into area equation to find maximum area.

Hope this helps! Let me know if anything doesn't make sense